Find the value of frac{d^{20}}{dx^{20}}(2cos | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Given expression is $y = 2\cos x \cos 3x$. Using the trigonometric identity $2\cos A \cos B = \cos(A+B) + \cos(A-B)$,we have: $y = \cos(3x+x) + \c

Vedclass · Accepted Answer

$2^{20}(\cos 2x + 2^{20}\cos 4x)$

Vedclass · Answer

(B) Given expression is $y = 2\cos x \cos 3x$. Using the trigonometric identity $2\cos A \cos B = \cos(A+B) + \cos(A-B)$,we have: $y = \cos(3x+x) + \cos(3x-x) = \cos 4x + \cos 2x$. Now,we need to find the $20^{th}$ derivative of $y$ with respect to $x$. Recall that $\frac{d^n}{dx^n}(\cos(ax)) = a^n \cos(ax + \frac{n\pi}{2})$. For $n=20$,$\frac{d^{20}}{dx^{20}}(\cos(ax)) = a^{20} \cos(ax + \frac{20\pi}{2}) = a^{20} \cos(ax + 10\pi) = a^{20} \cos(ax)$. Applying this to our expression: $\frac{d^{20}}{dx^{20}}(\cos 4x + \cos 2x) = 4^{20} \cos 4x + 2^{20} \cos 2x$. Since $4^{20} = (2^2)^{20} = 2^{40}$,the expression becomes $2^{40} \cos 4x + 2^{20} \cos 2x$. Factoring out $2^{20}$,we get $2^{20}(2^{20} \cos 4x + \cos 2x)$. This matches option $B$.

Find the value of $\frac{d^{20}}{dx^{20}}(2\cos x \cos 3x)$.

Similar Questions

If $f(x) = f'(x) + f''(x) + f'''(x) + \ldots$ and $f(0) = 1$,then $f(x)$ is equal to

If $y=e^{\sqrt{x \sqrt{x} \sqrt{x} \ldots}}, x>1$,then $\frac{d^2 y}{d x^2}$ at $x=\log _e 3$ is

The $n^{th}$ derivative of $x e^x$ vanishes when

If $y = \sin(\log_e x)$, then $x^2 \frac{d^2 y}{dx^2} + x \frac{dy}{dx}$ is equal to

If $f(x) = (x^2 - 1)^7$, then $f^{(14)}(x)$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module